(i) \(\begin{vmatrix}2 & 4\\2 & 1\end{vmatrix}\) = \(\begin{vmatrix}2x & 4\\6 & x\end{vmatrix}\)
⇒2 x 1 – 5 x 4 = 2x x x – 6 x 4)
⇒ 2- 20 = 2x2 – 24
⇒2x2 = 6
⇒ x2 = 3
⇒ x = ±√3
(ii) \(\begin{vmatrix}2 & 3\\4 & 5\end{vmatrix}\) = \(\begin{vmatrix}x & 3\\2x & 5\end{vmatrix}\)
⇒ 2 x 5 – 3 x 4 = x x 5 – 3 x 2x
⇒10 – 12 = 5x – 6x
⇒ -2 = -x
⇒ x = 2
In each of the following cases, state whether the function is one-one, onto or bijective. Justify your answer.
(i) f : R → R defined by f(x) = 3 – 4x
(ii) f : R → R defined by f(x) = 1 + x2
Show that the Modulus Function f : R → R, given by f(x) = |x|, is neither oneone nor onto, where | x | is x, if x is positive or 0 and |x| is – x, if x is negative.
Prove that the Greatest Integer Function f : R → R, given by f(x) = [x], is neither one-one nor onto, where [x] denotes the greatest integer less than or equal to x.
The total revenue in Rupees received from the sale of x units of a product is given by
R (x) = 13x2 + 26x + 15
Find the marginal revenue when x = 7.
Consider f : R → R given by f(x) = 4x + 3. Show that f is invertible. Find the inverse of f.
State with reason whether following functions have inverse
(i) f : {1, 2, 3, 4} → {10} with
f = {(1, 10), (2, 10), (3, 10), (4, 10)}
(ii) g : {5, 6, 7, 8} → {1, 2, 3, 4} with
g = {(5, 4), (6, 3), (7, 4), (8, 2)}
(iii) h : {2, 3, 4, 5} → {7, 9, 11, 13} with
h = {(2, 7), (3, 9), (4, 11), (5, 13)}
The vertices of ΔABC are A (3, 5, −4), B (−1, 1, 2), and C (−5, −5, −2).
Let f : X → Y be an invertible function. Show that f has unique inverse.
(Hint: suppose g1 and g2 are two inverses of f. Then for all y ∈ Y, fog1(y) = 1Y(y) = fog2(y). Use one-one ness of f).